Talk:Array of
What the order of solutions of legiattic array of marks: (a @ ( a @ ( a ... a @ ( a @ a) or conversely? Firstly, if it solves from right to left, then it seems to be dissimilar to standard array of mark. Secondly, why prime changes as b changes? For example, 3 @ X means {L,L,L,...,L,L,L}3,X. We require the prime to solve X. Might be better to define it as a @ b = {L,L,L,...,L,L,L}a,a? Ikosarakt1 (talk) 12:32, December 8, 2012 (UTC) Another notations It might be also interesting to define structures like: \(E(X)X\#\#\#X\) \(X \rightarrow X \rightarrow X \rightarrow X\) \(XX_X\) \(X \downarrow_{X} X\) For example, what \(25 \&\ 3\) means? Don't much known about it, but I know that this array, when fully expanded (without &'s), would have mega entries. Ikosarakt1 (talk ^ 22:29, March 26, 2013 (UTC) :\(X\) and \(\omega\) are one and the same. So if it works on \(\omega\), it works on \(X\). FB100Z • talk • 01:51, March 27, 2013 (UTC) :In response to specific questions, the down arrows seem to break down to the ups. Reaching for the nearest piece of scrap paper and doodling briefly I get \(\omega \downarrow^k \omega = \omega \uparrow\uparrow (k + 1)\). An \(X \downarrow_X X\) structure would then be a tetrational one. :Steinhaus-Moser notation also seems to degrade into up arrows. I get \(\text{Square}(\omega) = \epsilon_0\) and \(\text{Pentagon}(\omega) = \epsilon_1\). :Based on these I conclude that BEAF really is inescapable! FB100Z • talk • 01:59, March 27, 2013 (UTC) ::Wait a sec, it seems that \(\text{Pentagon}(\omega) = \epsilon_\omega\). Let me think about this :P FB100Z • talk • 02:02, March 27, 2013 (UTC) Even though \(X4 \&\ n \approx X \uparrow\uparrow X \&\ n\), we can't conclude that they are the same. The left expression should contains 256 entries, and right only 4. One of criteria to correctness of "array of" equations is the equivalent number of entries, and it can obtained when we evaluate each X into p and solve the array. So X can be considered as valiable, not an infinity, unlike \(\omega\). For example, \(X-1, {X \over 2}, \sqrt{(X)}\) are not equal to X, I believe. Ikosarakt1 (talk ^ 10:06, March 27, 2013 (UTC) :I'm still hanging on to omega here! In that case we need the surreal numbers. To get an X - 1 array, we need to position a pilot at \(\omega - 1\). To define structures such as \(\omega / 2\) and \(\sqrt{\omega}\), we have to face the general problem of continuizing BEAF. I don't think anyone's defined fundamental sequences for the surreals. Not yet, that is. FB100Z • talk • 18:46, March 27, 2013 (UTC) The way I like to think about X is like about a very big natural number. Actually, it must be larger than anything we will use (for example, TREE(3) should be large enough for most calculations). This allows us to make any arithmetical operations. And about surreals fundamental sequence - apart from doubly-limiting sequences used in definition of surreals, I guess there is no such thing. For example, \(\omega\) and \(\omega -\epsilon \) both have same lower limit sequences, only difference is upper limit. LittlePeng9 (talk) 19:44, March 27, 2013 (UTC) :The big problem is that surreals have no definition of a limit ordinal :P FB100Z • talk • 22:21, March 27, 2013 (UTC) I view X and \(\omega\) as different, or at least I interpret them differently. I interpret \(\omega\) as an ordinal to which we apply arithmetical operations that are defined by taking limits, whereas I interpret X as building abstract structures that are ordered by their growth rates as functions. So for example, \(\omega ^ {\omega \uparrow\uparrow \omega} = \omega ^ {\varepsilon_0} = \varepsilon_0 = \omega \uparrow\uparrow \omega\), but \(X ^ {X \uparrow \uparrow X} = X \uparrow\uparrow (X+1) > X \uparrow\uparrow X\). \(\omega \downarrow^k \omega \) goes up to \(\phi_{\omega}(0)\), but half as fast - \(\omega \downarrow^3 \omega = \varepsilon_0\), \(\omega \downarrow^5 \omega = \phi_2 (0) \), and so on. Pentagon\((\omega) = \phi_2 (0) \), and in general \(\omega\) in an n+3-gon is \(\phi_n(0)\). Deedlit11 (talk) 08:39, March 29, 2013 (UTC) Random section break So, is 33 & 3 = (3&3&3)&3 and 27 & 3 = (33) & 3? Jiawhein \(a\)\(l\) 05:15, April 14, 2013 (UTC) :No it's not. 33 & 3 = dimentri, and 27 & 3 = ultatri. 3&3&3&3 = (3&3&3)&3 is a whole larger number. — I want more 08:30, April 14, 2013 (UTC) :3 & 3 & 3 & 3 is the array with triakulus entries, which fill "triakularray"-space with 3's. When legiattic BEAF is used, it can be also expressed as {3,4 / 2}. Ikosarakt1 (talk ^ 09:05, April 14, 2013 (UTC) Just a quick question - isn't a&b={b,b,...,b} with a b's instead of b a's? Because for me 'a array of b' means array of b's of size a. LittlePeng9 (talk) 09:13, April 14, 2013 (UTC) :Yes, it is true (Bowers defined it so). Who defined that otherwise? Ikosarakt1 (talk ^ 09:17, April 14, 2013 (UTC) ::Article. Look at first sentence in "Usage" section. LittlePeng9 (talk) 09:26, April 14, 2013 (UTC) :::Thanks, I corrected that. Ikosarakt1 (talk ^ 09:33, April 14, 2013 (UTC) Add, multiply After so many years since this has been created, can anyone figure out how to work with additional and multiplicational arrays like \(1+2*3+5 \& 3\)? Ikosarakt1 (talk ^ ) 13:24, March 13, 2018 (UTC)